Probability puzzle for a fair die

Question

What is the expected number of times we have to roll a fair six-sided die to obtain that the number of times any two faces appear are equal (after the first rolling)?

Answer 1

These are just some back-of-the-envelope calculations, but I think they'll convince you that the probability that the event occurs is less than $1$, so that the expectation is infinite.

What is the probability that each number occurs $k$ times in $6k$ rolls? There are $\frac{(6k)!}{(k!)^6}$ acceptable sequences of rolls out of $6^{6k}$ possible sequences so that the probability is $$\frac1{6^{6k}}\frac{(6k)!}{(k!)^6}$$ Stirling's approximation gives $$\frac1{6^{6k}}\frac{(6k)!}{(k!)^6}\sim\frac1{6^{6k}}\frac{(6k)^{6k}e^{-6k}\sqrt{12\pi k}}{(k^ke^{-k}\sqrt{2\pi k})^6}=\frac{\sqrt3}4(\pi k)^{-5/2}$$

Numerical calculation shows that $$\sum_{k=1}^\infty\frac{\sqrt3}4(\pi k)^{-5/2}\approx\frac1{30}$$

an this sequence overstates the probability, since it would count sequence where the phenomenon more than once multiple times.

This should convince you that doing the calculations carefully, if it were thought worthwhile, would demonstrate that the phenomenon is not very likely to occur, and that the expected waiting time until it does is infinite.

Answer 2

Too long for a comment.

Staring from @saulspatz's answer, I wonder what it would become for a fair $n$-sided die and worked in the same spirit $$\frac1{n^{nk}}\frac{(nk)!}{(k!)^n}\sim \sqrt{n}\, (2 \pi )^{\frac{1-n}{2}} k^{\frac {1-n}2 }$$

$$P_n=\sum_{k=1}^\infty\frac1{n^{nk}}\frac{(nk)!}{(k!)^n}\sim \sqrt{n}\, (2 \pi )^{\frac{1-n}{2}} \, \zeta \left(\frac{n-1}{2}\right)$$ which is almost a straight line in a logarithmic scale.